Linear Algebra help..............?

Question

Here is a word problem and the questions that go with them I need any help possible with the questions.

A complany lease rental cars at 3 IL offices( Midway, O'Hare, and Loop) Records show that 60%% of the cars rented at Midway are returned to Midway and 20% and the other locations. 80% of the cars rented at OHare are returned to OHare and 10% to the other locations. 70% of the cars rented at the Loop are returned to the Loop, 10% returned to Midway, and 20% returned to OHare.
a) assuming that the info above describes a Markov chain write the transition matrix for this situation.
b) If a car is rented at Midway, what is the probability that it will be returned to the Loop after its second rental?
c) Over the long run, if all of the cars are returned, what proportion of the company's fleet will be located at each office?

Answer 1

a) Let a be the probability that a car is at Midway, b be the probability that the car is at O'Hare, and c be the probability that the car is at Loop. These may be summarized as the vector [a, b, c]. You know that after the transition matrix is applied to [1, 0, 0] (i.e. a car that is definitely rented from midway) will yield [.6, .2, .2]. Similarly, you know that T([0, 1, 0]) is [.1, .8, .1] and T([0, 0, 1]) is [.1, .2, .7]. So the transition matrix T must be:

[.6 .1 .1]
[.2 .8 .2]
[.2 .1 .7]

b) just compute T^2([1, 0, 0]). This gives you [.4, .32, .28], so the probability that it is at the loop after two rentals is .28

c) To find this you have to compute the limit, as n→∞, of T^n. This is best done by finding an eigenbasis of T- First we find the characteristic polynomial:

[.6-x .1 .1] \
[.2 .8-x .2] | - determinant of this
[.2 .1 .7-x] /

(.6-x) (.8-x) (.7-x) + .004 + .002 - .02 (.8-x) - .02 (.6-x) -.02 (.7-x)

-x³ + 2.1x² - 1.4x + 0.3

Setting this equal to zero, we find the roots of this polynomial, and therefore the eigenvalues of the matrix are:

x=1, x=.5, x=.6

To find the corresponding eigenvectors, we find the kernel of T-xI, where I is the identity matrix, and x is one of the eigenvalues. For x=1, we find the kernel of:

[-.4 .1 .1]
[.2 -.2 .2]
[.2 .1 -.3]

The rref of this is:

[1 0 -2/3]
[0 1 -5/3]
[0 0 0]

So the eigenvector for the eigenvalue 1 is [2/3, 5/3, 1] (or any scalar multiple of that vector). Similar methods allow you to find the two other eigenvectors [1, 0, -1] and [0, 1, -1]. This gives you the eigenbasis matrix S:

[2/3 1 0]
[5/3 0 1]
[3/3 -1 -1]

Compute its inverse, S^-1:

[.3 .3 .3]
[.8 -.2 -.2]
[-.5 .5 -.5]

And thus you can write the original matrix T as SMS^-1, where M is:

[1 0 0]
[0 .5 0]
[0 0 .6]

Multiply it out just to check your work. Now that we have diagonalized T, it is easy to find large powers of it, since for any 3 by 3 diagonal matrix D having diagonal elements (D_1, D_2, D_3), D^n is the diagonal matrix with elements (D_1^n, D_2^n, D_3^n). And T^n = SMS^-1 SMS^-1 SMS^-1... SMS^-1. Canceling the S^-1 S pairs, we have T^n = SM^nS^-1, and computing M^n is easy. Taking the limit as n→∞, we have M^n =

[1 0 0]
[0 0 0]
[0 0 0]

So multiplying our final product, we have:

[.2 .2. 2]
[.5 .5 .5]
[.3 .3 .3]

Thus, regardless of where the car starts out, over the long run, it will have a 20% chance of ending up at Midway, 50% chance of ending up at O'Hare, and a 30% chance of ending up at Loop.

And yes, this was a long problem. Most of the work was done in diagonalizing the matrix T, so if you have access to a computer algebra system, and your teacher has low requirements for showing work, I highly recommend that you let it do all the tedious work of diagonalizing the matrix.

Answer 2

There are no prerequisites really: linear algebra is pretty self contained.